Question

It is desired to construct a solenoid that will have a resistance of 4.00 Ω (at 20°C) and produce a magnetic field of 4.00 ✕ 10⁻² T at its center when it carries a current of 4.25 A. The solenoid is to be constructed from copper wire having a diameter of 0.500 mm. If the radius of the solenoid is to be 1.00 cm, determine the following. (The resistivity of copper at 20°C is 1.7 ✕ 10⁻⁸ Ω · m.)

(a) the number of turns of wire needed to build the solenoid

(b) the length the solenoid should have

Answer 1

R = 4 ohm

B = 4.00 ✕ 10⁻² T

I = 4.25 A

diameter of wire(d) = 0.500 mm

radius of the solenoid is r = 1.00 cm = 1 X 10^-2 m

Area of solenoid A = 3.14 (1 X 10^-2 )²

A = 3.14 X 10^-4 m²

a) The magnetic field in a solenoid is given by

B = u_o nI

4.00 ✕ 10⁻² = 4pi X 10^-7 X n X 4.25

n = 7.45 X 10³ turns /m

Let us calculate the total length of solenoid

R = pl/A

l = RA/p

A is the area of cross section of wire

A = pi(d/2)²

A = 3.14 (0.5 X 10^-3/2)²

A = 1.96 X 10^-7 m²

l = RA/p

l = 4 (1.96 X 10^-7) / (1.7 ✕ 10⁻⁸)

l = 46.12 m

The length of solenoid = 46.12 m

Total no. of turns N = nl = 7.45 X 10³ turns /m X 46.12 = 3.44 X 10⁵